Problem: Simplify the following expression: $\dfrac{7q^2}{63q^2}$ You can assume $q \neq 0$.
Solution: $ \dfrac{7q^2}{63q^2} = \dfrac{7}{63} \cdot \dfrac{q^2}{q^2} $ To simplify $\frac{7}{63}$ , find the greatest common factor (GCD) of $7$ and $63$ $7 = 7$ $63 = 3 \cdot 3 \cdot 7$ $ \mbox{GCD}(7, 63) = 7 $ $ \dfrac{7}{63} \cdot \dfrac{q^2}{q^2} = \dfrac{7 \cdot 1}{7 \cdot 9} \cdot \dfrac{q^2}{q^2} $ $\phantom{ \dfrac{7}{63} \cdot \dfrac{2}{2}} = \dfrac{1}{9} \cdot \dfrac{q^2}{q^2} $ $ \dfrac{q^2}{q^2} = \dfrac{q \cdot q}{q \cdot q} = 1 $ $ \dfrac{1}{9} \cdot 1 = \dfrac{1}{9} $